The degree three polynomial f(x) with real coefficients and
leading coefficient 1, has 4 and 3+i among its roots. 
Express f(x) as a product of linear and quadratic polynomials
polynomials with real coefficients. This question I'm having 
trouble with but I came up witha an answer of:
f(x)=(x-4)(x^2-6x-9) is this right. thanks for looking at 
this question.

No that's not correct.  You must know the following facts about
polynomials:

1. A degree n polynomial has n roots, counting multiplicities.
2. If r is a root of a polynomial then (x-r) is a factor of the
   polynomial.
3. If a polynomial with real coefficients has p+qi as one root,
   it also has its conjugate p-qi as another root.

A, By 2 above, since 4 is a root then (x-4) is a factor.
         (You did this)
B. By 2 above, since 3+i is a root then [x-(3+i)] is a factor
   of the polynomial. 
C. By 3 above, since 3+i is a root then 3-1 is a factor of the
   polynomial.
D. By C and 2, since 3-i is a root then [x-(3-i)] is a factor 
   of the polynomial.
E. By 1, there are no more roots besides 4, 3+i, and 3-1

Therefore f(x) is the product of the three factors. It will 
have leading coefficient 1 because the c oefficients of x in 
all three factors is 1.

[Note: If it had had a leading coefficient other than 1, we 
 would have to multiply the polynomial by it, too, but thet 
 is unnecessary here.]

So we have:

f(x) = (x - 4)[x - (3+i)][x - (3-i)]

But the two factors containing imaginary numbers have to be
multiplied together, since all coefficients must be real.
We can use FOIL to multiply the last two factors together:

f(x) = (x - 4)[x - (3+i)][x - (3-i)]
f(x) = (x - 4)[x² - (3-i)x - (3+i)x + (3+i)(3-i)]
f(x) = (x - 4)[x² - (3x-ix) - (3x+ix) + (9-3i+3i-i²)]
f(x) = (x - 4)[x² - 3x + ix - 3x - ix + (9-i²)]
f(x) = (x - 4)[x² - 6x + 9 - i²]

Now since i² = -1 we substitute (-1) for i².

f(x) = (x - 4)[x² - 6x + 9 - (-1)]
f(x) = (x - 4)[x² - 6x + 9 + 1]
f(x) = (x - 4)[x² - 6x + 10]
f(x) = (x - 4)(x² - 6x + 10)

Edwin
AnlytcPhil@aol.com